Question

guys please please please please solve it. its urgent​

Answer 1

Answer:

20

I hope it'll be helpful

Answer 2

Given:

In Δ ABC,

∠ACD = 40°

AD = CD

AB = BD

To find:

∠ABC

Solution:

We have,

AD = CD

∴ ∠ACD = ∠DAC = 40° .... [∵ angles opposite to equal sides are also equal]

In Δ ADC, we have

∠ACD + ∠DAC + ∠ADC = 180° ..... [Angle sum property of a triangle]

⇒ 40° + 40° + ∠ADC = 180°

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°

Also,

∠ADC + ∠ADB = 180° ..... [Linear Pair]

⇒ 100° + ∠ADB = 180°

⇒  ∠ADB = 180° - 100°

⇒ ∠ADB = 80°

We are given that,

AD = BD

∴ ∠ADB = ∠BAD = 80° .... [∵ angles opposite to equal sides are also equal]

Now,

In Δ ABD, we have

∠ADB + ∠BAD + ∠ABC = 180° ..... [Angle sum property of a triangle]

⇒ 80° + 80° + ∠ABC = 180°

⇒ 160° + ∠ABC = 180°

⇒ ∠ABC = 180° - 160°

⇒ ∠ABC = 20°

Thus, $\boxed{\bold{\angle ABC = \underline{20\° }}}.$

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